Let [tex]A(t)[/tex] be the amount of salt (in pounds) in the tank at time [tex]t[/tex]. We're given that[tex]A(0)=10\text{ lb}[/tex].
The rate at which the amount of salt in the tank changes is given by the ODE
[tex]A'(t)=\dfrac{2\text{ gal}}{1\text{ min}}\cdot\dfrac{0\text{ lb}}{1\text{ gal}}-\dfrac{1\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{50+(2-1)t\text{ gal}}[/tex]
[tex]A'(t)+\dfrac{A(t)}{50+t}=0[/tex]
[tex](50+t)A'(t)+A(t)=0[/tex]
[tex]\bigg((50+t)A(t)\bigg)'=0[/tex]
[tex](50+t)A(t)=C[/tex]
[tex]A(t)=\dfrac C{50+t}[/tex]
Given that [tex]A(0)=10[/tex], we find that
[tex]10=\dfrac C{50+0}\implies C=500[/tex]
so that the amount of salt in the tank is described by
[tex]A(t)=\dfrac{500}{50+t}[/tex]
The tank will be filled when [tex]50+t=100[/tex], or after [tex]t=50[/tex] minutes. At this time, the amount of salt in the tank is
[tex]A(50)=\dfrac{500}{50+50}=5\text{ lb}[/tex]
